RT = J where R = rate, T = time and J = job(s)
Sarah's rate (RS) can be calculated since we know her time (TS) to complete the task is 5 hours and there is one job (or task) so J is 1.
RS(5) = 1.
RS(5) = 1 Dividing both sides by 5 we get
5 5
RS = 1/5
Ben's rate (RB) can be calculated since we know his time (TB) to complete the task is 4 hours and there is one job (or task) so J is 1.
RB(4) = 1.
RB(4) = 1 Dividing both sides by 4 we get
4 4
RB = 1/4
RB(4) = 1 Dividing both sides by 4 we get
4 4
RB = 1/4
Since they are both working together the formula will be...
RSTS + RBTB = J
We also know Ben worked 2 hours so TB is 2. Substituting we get:
(1/5)TS + (1/4)(2) = (1)
(1/5)TS + 1/2 = 1 [Multiplying (1/4)(2)]
(1/5)TS = 1/2 [Subtracting 1/2 from both sides]
TS = 5/2 or 2.5 [Multiplying both sides by 5]
This is reasonable since Ben is able to do the entire task in 4 hours he should be able to do half the task in 2 hours. That leaves half the task for Sarah to do, and since she can do the whole task by herself in 5 hours she should be able do her half in 2.5 hours.
To answer the question, Sarah must work an extra 1/2 (or 0.5) hour(s) to complete the task (5/2 - 2 = 1/2 -or- 2.5 - 2 = 0.5).
Steven B.
tutor
Same formula. Start with:
RSTS + RBTB = J
(1/5)TS + (1/4)TB = 1 Substitute for rates and job.
Note that since they both start and stop at the same time, TS = TB, so substituting we get:
(1/5)TS + (1/4)(TS) = 1
At this point there are several approaches to finding the answer but since I am not fond of dealing with fraction, I can multiply all terms by the common denominator (20) to get rid of them.
20[(1/5)TS + (1/4)(TS]) = 20(1)
4TS + 5TS = 20 Distribute (and simplify)
9TS = 20 Adding (since they are both TS)
Then divide both sides by 9
9TS = 20
9 9
TS = 20/9 and since TB = TS, TB = 20/9
20/9 = 22.2222222...
The basic formula works for a lot things, and you'll notice a change in variable name only. If you are doing distance (RT = D, rate x time = distance) or mixture (PQ = S, where P = %, Q = Quantity and S = Solution, although this one is a little more complicated than that). Often you will need to write 2 (or more) equation that can then be manipulated and combined (usually by substitution) to arrive at one equation in one variable (for example, in the 2nd problem, the 1st two equations are added together and then we had to write a third equation TB = TS in order to substitute and arrive at a new equation with just one variable in it).
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04/24/14
Steven B.
tutor
I wrote an answer to your second question but apparently it didn't save correctly. I'll type it in again tomorrow. WAY past my bedtime now.
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04/24/14
Steven B.
tutor
You use the same formula [In fact this formula is similar to the "other" distance formula RT = D where R = Rate, T = Time and D = Distance. And the mixture formula P1Q1 + P2Q2 = P3(Q1 + Q2)]. Anyway...
RSTS + RBTB = J Starting from here...
(1/5)TS + (1/4)TB = (1) Now we have a problem, because we have one equation with 2 variables in it
If only I could find another equation! Ah Ha! Since they both worked the same amount of time, TS = TB.
(1/5)TS + (1/4)TS = (1) Substituting TS for TB.
At this point there are several approaches one can use to simplify the equation, but I don't like to have to deal with fractions, so I am going to multiply by the LCD (least common denominator), which is 20, to get rid of them.
20[(1/5)TS + (1/4)TS] = 20(1) x20 which gives us...
4TS + 5TS = 20 Next we add the like terms...
9TS = 20 Now dividing both sides by 9
TS = 20/9 = 2 & 2/9 hrs. or 2.2222... hrs. or ≈ 2.22 hrs. (or 2 hour 13 & 1/3... min., etc. etc. etc.)
And finally if TS = 20/9 then TB = 20/9
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04/24/14
Karen K.
04/21/14